Question 172786

First let's graph {{{x-4y=-12}}}. 



In order to graph {{{x-4y=-12}}}, we need to solve for "y"



{{{x-4y=-12}}} Start with the given equation.



{{{-4y=-12-x}}} Subtract {{{x}}} from both sides.



{{{-4y=-x-12}}} Rearrange the terms.



{{{y=(-x-12)/(-4)}}} Divide both sides by {{{-4}}} to isolate y.



{{{y=((-1)/(-4))x+(-12)/(-4)}}} Break up the fraction.



{{{y=(1/4)x+3}}} Reduce.





Looking at {{{y=(1/4)x+3}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=1/4}}} and the y-intercept is {{{b=3}}} 



Since {{{b=3}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,3\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,3\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15))
)}}}



Now since the slope is comprised of the "rise" over the "run" this means

{{{slope=rise/run}}}


Also, because the slope is {{{1/4}}}, this means:


{{{rise/run=1/4}}}



which shows us that the rise is 1 and the run is 4. This means that to go from point to point, we can go up 1  and over 4




So starting at *[Tex \LARGE \left(0,3\right)], go up 1 unit 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15)),
  blue(arc(0,3+(1/2),2,1,90,270))
)}}}


and to the right 4 units to get to the next point *[Tex \LARGE \left(4,4\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15)),
  blue(circle(4,4,.15,1.5)),
  blue(circle(4,4,.1,1.5)),
  blue(arc(0,3+(1/2),2,1,90,270)),
  blue(arc((4/2),4,4,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(1/4)x+3}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,(1/4)x+3),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15)),
  blue(circle(4,4,.15,1.5)),
  blue(circle(4,4,.1,1.5)),
  blue(arc(0,3+(1/2),2,1,90,270)),
  blue(arc((4/2),4,4,2, 180,360))
)}}} So this is the graph of {{{y=(1/4)x+3}}} through the points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(4,4\right)]



--------------------



Now let's graph {{{y=-4+2x}}}



{{{y=-4+2x}}} Start with the given equation



{{{y=2x-4}}} Rearrange the terms.





Looking at {{{y=2x-4}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=2}}} and the y-intercept is {{{b=-4}}} 



Since {{{b=-4}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-4\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,-4\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-4,.1)),
  blue(circle(0,-4,.12)),
  blue(circle(0,-4,.15))
)}}}



Now since the slope is comprised of the "rise" over the "run" this means

{{{slope=rise/run}}}


Also, because the slope is {{{2}}}, this means:


{{{rise/run=2/1}}}



which shows us that the rise is 2 and the run is 1. This means that to go from point to point, we can go up 2  and over 1




So starting at *[Tex \LARGE \left(0,-4\right)], go up 2 units 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-4,.1)),
  blue(circle(0,-4,.12)),
  blue(circle(0,-4,.15)),
  blue(arc(0,-4+(2/2),2,2,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,-2\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-4,.1)),
  blue(circle(0,-4,.12)),
  blue(circle(0,-4,.15)),
  blue(circle(1,-2,.15,1.5)),
  blue(circle(1,-2,.1,1.5)),
  blue(arc(0,-4+(2/2),2,2,90,270)),
  blue(arc((1/2),-2,1,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=2x-4}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,2x-4),
  blue(circle(0,-4,.1)),
  blue(circle(0,-4,.12)),
  blue(circle(0,-4,.15)),
  blue(circle(1,-2,.15,1.5)),
  blue(circle(1,-2,.1,1.5)),
  blue(arc(0,-4+(2/2),2,2,90,270)),
  blue(arc((1/2),-2,1,2, 180,360))
)}}} So this is the graph of {{{y=2x-4}}} through the points *[Tex \LARGE \left(0,-4\right)] and *[Tex \LARGE \left(1,-2\right)]



--------------------------------------------------



Now let's graph the two equations on the same coordinate system:


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,(1/4)x+3,2x-4)
)}}} Graph of {{{y=(1/4)x+3}}} (red) and graph of {{{y=2x-4}}} (green)


From the graph, we can see that the two lines intersect at the point (4,4). So the solution is (4,4)